Answer:
Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>
Part b: <em>The case in such a way that the chances are maximized so the case where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>
Part c: <em>The minimum and maximum probabilities of winning for n number of balls are such that </em>
- <em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>
- <em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>
Step-by-step explanation:
Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as
P(A)=P(A')=0.5
Now the probability of finding the black ball is given as
P(B)=P(B∩A)+P(P(B∩A')
P(B)=(P(B|A)P(A))+(P(B|A')P(A'))
Now there can be four cases as follows
Case 1: When all the four balls are in urn A and no ball is in urn A'
so
P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as
P(B)=(0.25*0.5)+(0*0.5)
P(B)=0.125;
Case 2: When the black ball is in urn A and 3 white balls are in urn A'
so
P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as
P(B)=(1*0.5)+(0*0.5)
P(B)=0.5;
Case 3: When there is 1 black ball and 1 white ball in urn A and 2 white balls are in urn A'
so
P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as
P(B)=(0.5*0.5)+(0*0.5)
P(B)=0.25;
Case 4: When there is 1 black ball and 2 white balls in urn A and 1 white ball are in urn A'
so
P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as
P(B)=(0.33*0.5)+(0*0.5)
P(B)=0.165;
Part a:
<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>
Part b:
<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>
Part c:
The minimum and maximum probabilities of winning for n number of balls are such that
- when all the n balls are placed in one of the urns the probability of the winning will be least given as
P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as
P(B)=(1/n*1/2)+(0*0.5)
P(B)=1/2n;
- when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as
P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as
P(B)=(1/1*1/2)+(0*0.5)
P(B)=0.5;
